SOLUTION: You need a 40%saline solution for an experiment, but only have 25% saline and 75% saline solutions. How much of each would you have to combine to create 50ml of 40% saline solutio

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Question 184451: You need a 40%saline solution for an experiment, but only have 25% saline and 75% saline solutions. How much of each would you have to combine to create 50ml of 40% saline solution?
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
You need a 40%saline solution for an experiment, but only have 25% saline and 75% saline solutions. How much of each would you have to combine to create 50ml of 40% saline solution?
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Equation:
salt + salt = salt
0.25x + 0.75(50-x) = 0.40*50
Multiply thru by 100 to get:
25x + 75*50 - 75x = 40*50
-50x = -35*50
x = 35mL (amt. of 25% solution in the mixture)
50-x = 15mL (amt. of 75% solution in the mixture)
======================================================
Cheers
Stan H.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let a = ml of 25% solution needed
Let b = ml of 75% solution needed
In words:
((salt in 25% solution) + (salt in 75% solution)) / (total amount of solution) =
40%
given:
salt in 25% solution = .25a ml
salt in 75% solution = .75b ml
Total amount of solution = 50 ml
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(1) a+%2B+b+=+50
(2) %28.25a+%2B+.75b%29+%2F+50+=+.40
(2) .25a+%2B+.75b+=+20
Multiply both sides of (1) by .25
(1) .25a+%2B+.25b+=+12.5
Subtract (1) from (2)
(2) .25a+%2B+.75b+=+20
(1) -.25a+-+.25b+=+-12.5
(3) .50b+=+7.5
(3) b+=+15
and
(1) a+%2B+b+=+50
(1) a+%2B+15+=+50
a+=+35
35 ml of 25% solution and 15 ml of 75% solution are needed
check:
(2) %28.25a+%2B+.75b%29+%2F+50+=+.40
(2) %28.25%2A35+%2B+.75%2A15%29+%2F+50+=+.40
(2) %288.75+%2B+11.25%29+%2F+50+=+.40
(2) +20+%2F+50+=+.40
(2) .40+=+.40
OK